A rectangular room is 6 meters longer than it is wide, and its perimeter is 24 meters. Find the dimension of the room.

The length is :   meters and the width is  meters.


To solve the problem, let's define the dimensions of the room:

- Let \( W \) be the width of the room in meters.
- Since the room is 6 meters longer than it is wide, the length \( L \) of the room is \( W + 6 \) meters.

The perimeter \( P \) of a rectangle is given by:
\[
P = 2L + 2W
\]
We are told that the perimeter is 24 meters, so:
\[
2(W + 6) + 2W = 24
\]

Now, let's solve for \( W \):

1. Expand the equation:
   \[
   2W + 12 + 2W = 24
   \]

2. Combine like terms:
   \[
   4W + 12 = 24
   \]

3. Subtract 12 from both sides:
   \[
   4W = 12
   \]

4. Divide by 4:
   \[
   W = 3 \text{ meters}
   \]

Now, to find the length \( L \):

\[
L = W + 6 = 3 + 6 = 9 \text{ meters}
\]

So, the dimensions of the room are:
- Length: \( 9 \) meters
- Width: \( 3 \) meters

Would you like further explanation or have any questions?

Here are some related questions to consider:

1. How would the perimeter change if the room was 8 meters longer than it is wide?
2. How do you find the area of the room given these dimensions?
3. What if the perimeter was 30 meters instead of 24 meters? How would the dimensions change?
4. How can you set up an equation if the length is twice the width?
5. What if the length was \( W + 10 \) meters? How would you find the dimensions?

**Tip:** When solving problems with relationships between dimensions and perimeter, setting up equations based on these relationships is key to finding the solution.

Learn how to find the dimensions of a rectangular room when given its perimeter and the relationship between its length and width. This problem involves solving linear equations and understanding geometric properties. Suitable for students in Grades 7-9.

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